\section{Flat pricing market}
\label{sec:flat}

\subsection{Market Model}
We first consider a market model under the flat pricing scheme. 
In this market model, the operator decides on the price vector $\bm{p}=(p_l:l \in
\{ m, o, c\})$ in order to maximize the revenue $R$ by solving the following
problem:
\begin{eqnarray}
\label{eq:provider}
{\bf Provider:}&& \qquad \max_{p_{m},p_{o},p_{c} \ge 0}R. 
\end{eqnarray}
In the flat pricing scheme, the revenue in (\ref{eq:revenue}) is simplified to
\begin{equation}
\label{eq:revenueflat}
 R=N \Big(\sum_{l} p_l \alpha_l - c_f \cdot (\alpha_o+\alpha_c)\Big).
\end{equation}

The subscription ratios ${\bm \alpha}$ vary by price. 
We assume that users are selfish and try to maximize individual (expected) utility.
Thus, a user of type
$\gamma$ selects the service $l^*(\gamma)$ that maximizes his or her
net-utility: 
\begin{eqnarray}
\label{eq:user_flat}
{\bf User:} && l^*(\gamma) =\arg \max_{l \in \{ m,o,c \}} \tilde{U}_{l}({\bm{x}};\gamma),
\end{eqnarray}
%
when his or her maximum net-utility is {\em positive,} and he or she does not select
any service, otherwise.  This market model can be modeled as a two-stage
sequential game, where the operator determines the price vector to
maximize the revenue in the first stage, and then, users select one of the
services (or exit from the market) according to the price vector provided by the
operator.

%Solving the two-stage game is non-trivial due to inter-coupling between $\bm{\alpha}$
%and $\bm{p}$. Typically, we use backward induction to solve the sequential game, i.e,
%for given $\bm{p}$, we solve the user's problem to find $\bm{\alpha}$. Then, we optimize
%the first stage game to decide the price.  However, this problem requires
%a Newton-type backward induction as it is a fixed point problem.
%Observe that even for a given $\bm{p},$ computing  $\bm{\alpha}$ is non-trivial because
%the net utility of each user affects $\bm{\alpha}$ due to the problem
%{\bf User}, and also $\bm{\alpha}$ affects the net utility due to the
%expected throughput's dependence of $\bm{\alpha}.$  To show the dependency,
%we use $x_j(\bm{\alpha}), j \in \{m,o,c\}.$

%The following theorem enables us to solve the problem very efficiently.
%Let us define $\set{A}$ given by:
%\[ \set{A} = \{ \bm{\alpha}| x_m(\bm{\alpha}) \leq x_o(\bm{\alpha}) \leq
%x_c(\bm{\alpha}) \}. \]
%\begin{theorem}
%\label{thm:palpha}
%{\it For all $\bm{\alpha} \in \set{A},$ the optimal revenues of the provider
%in (\ref{eq:provider}) are the same. }
%\end{theorem}

%We omit the proof due to page limitation but we provide the sketch of
%proof. Note that for a given $\bm{\alpha}$, there may exist multiple
%price vectors that leads to the $\bm{\alpha}.$ Recall that if a price
%vector is given, users selects their service to maximize their own
%utility (i.e., by solving {\bf User}), which results in some ${\bm{\alpha}}.$ Since the revenue is
%computed as $N(\alpha_{m}p_{m}+\alpha_{o}p_{o}+\alpha_{c}p_{c}),$ the
%prices for service type $j \in \{m,o,c \}$ where $\alpha_{j}=0$ do not
%contribute to the revenue. Thus, it suffices to consider only when $p_m,
%p_o, p_c >0,$ in which case, if $\bm{\alpha} \in \set{A},$ we can prove
%that the resulting revenue is uniquely determined. 
%Denote by $\set{P}(\bm{\alpha})$ the set of all $\bm{p}$ that lead to
%the $\bm{\alpha}$ in the way as described above.  % Thus, when the operator gives services with any
% $\bm{p}\in \set{P}(\bm{\alpha})$, the distribution of users' service
% type becomes the given $\bm{\alpha}$ as each user selects his service
% based on the problem $\bf User$.

%Since, from Theorem~\ref{thm:palpha}, it suffices to use any $\bm{p}$ in the
%set $\set{P}(\bm{\alpha})$, we will use a  $\bm{p^{*}} \in \set{P}(\bm{\alpha^{*}})$, which satisfies
%\begin{eqnarray}
%p_{m}^{*} & = & (1-\alpha_{m}-\alpha_{o}-\alpha_{c})g_{m}(x),\\
%p_{o}^{*} & = & (1-\alpha_{o}-\alpha_{c})(g_{o}(x)-g_{m}(x))+p_{m}^{*},\\
%p_{c}^{*} & = & (1-\alpha_{c})(g_{c}(x)-g_{o}(x))+p_{o}^{*},\end{eqnarray}
%for the optimal price. Therefore, we can have the closed form of the revenue expressed only by $\bm{\alpha}.$

\subsection{Traffic}

We assume that users are saturated and have sufficient data to
transmit whenever possible. The (average) amount of data generated by each user
depends on service type, capacities $C_M,C_O,$ and $C_C,$ and the
scheduling discipline of BSs for competitive users.  In particular, we
simply assume that a BS serves 
its served users equally~\footnote{in view of long term average, users
  have the same data rate when their mobility patterns are homogeneous.}. Under this
fairness assumption, the average service rate of a user served by macro
BSs is inversely proportional to the number of users in a macro BSs,
given by
\begin{eqnarray}
  \label{eq:xm}
  x^{M}&=&C_{M}/(1+\sum_{l\in\{m,o,c\}}\pi_{l}^{M}\alpha_{l}N),
\end{eqnarray}
where the denominator in (\ref{eq:xm}) corresponds to the total number
of users in a macro BS, whereas a user is in the macro BS's
service. Similarly, the service rates for users
with open and closed femto BSs are given by
% Since we consider a homogeneous situation in terms of users and
% femtocells, each open femto AP has the same number of users
% expectedly. Thus, the service rate of a user for open femto APs is
\begin{eqnarray}
x^{O}  &=& 
C_{O}/(1+\sum_{l\in\{m,o,c\}}\pi_{l}^{O}\alpha_{l}N/\alpha_{o}N), \label{eq:xo}
\\
x^{C} & = & C_{C} \label{eq:xc},
\end{eqnarray}
respectively, where note that $\sum_{l\in\{m,o,c\}} \pi_{l}^{O}\alpha_{l}N/\alpha_{o}N$ is the average number of
users visiting an open femto BS.

\subsection{Equilibrium}

\begin{figure}[t!]
   \begin{center}
     \includegraphics[width=0.95\columnwidth]{fig/net_price_flat_volume.eps}
   \end{center}
   \caption{Examples of users' net-utility for each service type: (a)
     flat pricing and (b) partial volume pricing}
   \label{fig:flat_price}
\end{figure}

As shown in Fig.~\ref{fig:flat_price}(a),
user subscription ratios
$\bm{\alpha} = (\alpha_l: l \in \{ m, o, c\} )$ are a function of
price level $\bm{p}$ and data rate $\bm{x}$. Let $\gamma_{i}$ be a
point $\tilde{U}_{i}(x;\gamma_{i}) = 0$ for all $i
\in \{m,o,c\}$ and $\gamma_{ij}$ be a point such that 
$\tilde{U}_{i}(x;\gamma_{ij}) = \tilde{U}_{j}(x;\gamma_{ij})$ for all $i,j
\in \{m,o,c\}$.
% Then, we can characterize $\bm{\alpha}$ as follows:
% \begin{eqnarray*}
%   \alpha_{m}=\frac{\gamma_{mo}-\gamma_{m}}{\bar{\gamma}}, \quad
%   \alpha_{o}=\frac{\gamma_{oc}-\gamma_{mo}}{\bar{\gamma}}, \quad
%   \alpha_{c}=\frac{\bar{\gamma}-\gamma_{oc}}{\bar{\gamma}}.
% \end{eqnarray*}


Finding the equilibrium of the flat pricing market is difficult
because of the complex inter-play between $\bm{\alpha}$ and $\bm{p}$.
We typically use a backward induction to solve the sequential game,
that is, for a given $\bm{p}$, we solve the user's problem to find the
corresponding $\bm{\alpha}(\bm{p})$. Then, we optimize the first stage
game to decide on the equilibrium price $\bm{p}^\star$ by solving
$\max_{\bm{p}} R(\bm{p})$ from (\ref{eq:revenueflat}).  However, our
problem requires a Newton-type backward induction to numerically solve
a complex fixed point problem because of the lack of a closed form (see the outer loop in Fig.~\ref{fig:eq_diff}). 

\begin{figure}[t!]
  \begin{center}
    \includegraphics*[width=0.7\columnwidth]{fig/eq_diff2.eps}
  \end{center}
  \caption{Flat pricing: Coupling in the two-stage sequential game}
  \label{fig:eq_diff}
\end{figure}

Note that even for a given $\bm{p},$ computing $\bm{\alpha}$ by solving
the problem $\mathbf{User}$ is difficult. This is because the
net-utility of each user affects $\bm{\alpha}$ because of the problem
${\bf User}$, and also because $\bm{\alpha}$ affects the net-utility
owing to the achieved data rates' dependence on $\bm{\alpha}.$ In
order to explicitly represent this dependency, we denote
$\bm{x}(\bm{\alpha})= (x^{L}(\bm{\alpha}): L \in \{M,O,C\})$ for a
given $\bm{\alpha}.$

Theorem~\ref{thm:palpha} enables us to compute the equilibrium
efficiently (the proof is presented in Appendix).  The basic idea is
that we express revenue $R=R(\bm{\alpha})$ as a function
$\bm{\alpha}$, not $\bm{p}$ by finding $\bm{p}$'s closed form
w.r.t. $\bm{\alpha}.$ Then, revenue $R(\bm{\alpha})$ can be simply
maximized. Note that for a given $\bm{\alpha},$ there may exist
multiple values of $\bm{p}$ that lead to the same subscription ratio
$\bm{\alpha}.$ The set of all such $\bm{p}$ is denoted by
$\set{P}(\bm{\alpha}).$ Theorem~\ref{thm:palpha} also states that it
is sufficient to consider {\em one} $\bm{p} \in \set{P}(\bm{\alpha})$
because all price vectors in $\set{P}(\bm{\alpha})$ lead the same.



%   Define $\set{A}$ by a set of $\bm{\alpha}$, such that the data rates
%   over macro, open-femto, and closed-femto BSs are ordered:
% \begin{equation} \set{A} \triangleq \{ \bm{\alpha} \mid x^{M}(\bm{\alpha}) \leq
% x^{O}(\bm{\alpha}) \leq x^{C}(\bm{\alpha}) \}. \end{equation}
 
% For ease of presentation of Theorem~\ref{thm:palpha}, we will use the
% index variables $I,$ $J$, and $K$ to distinctly refer to one of the {\em
%   BS types} $M,O,$ and $C.$ Similarly, the index variables $i,$ $j$, and
% $k$ are used to refer to one of the {\em service types} $m,$ $o,$ and
% $c,$ following the same sequence of $I,$ $J,$ and $K.$ For example, when
% $I=\singleq{M}$, $J=\singleq{C},$ $K=\singleq{O},$ then $i=\singleq{m},$
% $j=\singleq{c},$ and $k=\singleq{o}.$

For ease of presentation of Theorem~\ref{thm:palpha}, we will use the
index variables $i,$ $j$, and $k$ to distinctly refer to one of the
{\em service types} $m,$ $o,$ and $c.$

\begin{theorem}
  \label{thm:palpha} Let $T_l :=\sum_{L \in
  \{M,O,C \}} (x^L)^\theta \pi_l^L,$ $l \in \{m,o,c \}.$ Consider the following set $\set{A}$:
  \begin{eqnarray*}
    \set{A} \triangleq \{ \bm{\alpha} \mid T_i \leq T_j \leq T_k \}. 
  \end{eqnarray*}
  For any $\bm{\alpha} \in \set{A,}$ 
  % Consider $\bm{\alpha}$ in the set $\{ \bm{\alpha} \mid x^{I}(\bm{\alpha}) \leq
  % x^{J}(\bm{\alpha}) \leq x^{K}(\bm{\alpha}) \}.$ % Then, 
  %  Let, for a $\bm{\alpha}$, $\{ x^{I}(\bm{\alpha}) \leq
  % x^{J}(\bm{\alpha}) \leq x^{K}(\bm{\alpha}) \},$ where $\{I, J,K \} =
  % \{ M , O , C \},$ and service type $i$, $j$, and $k$ follow the same
  % character mapping between $\{I, J,K \}$ and $\{ M , O , C \}$.
  there exists a $\bm{p}'=(p'_m,p'_o,p'_c) \in
  \set{P}(\bm{\alpha}),$ such that
\begin{eqnarray}
  \label{eq:thmpalpha}
p_{i}' & = &U_{i}(\bm{x}(\bm{\alpha});\gamma_{i}),\cr
p_{j}' & = & U_{j}(\bm{x}(\bm{\alpha});\gamma_{ij})-U_{i}(\bm{x}(\bm{\alpha});\gamma_{ij})+p_{i}',\cr
p_{k}' & = & U_{k}(\bm{x}(\bm{\alpha});\gamma_{jk})-U_{j}(\bm{x}(\bm{\alpha});\gamma_{jk})+p_{j}',
\end{eqnarray}
where $\gamma_{i}=1-\alpha_{i}-\alpha_{j}-\alpha_{k}$,
$\gamma_{ij}=1-\alpha_{j}-\alpha_{k},$ and $\gamma_{jk}=1-\alpha_{k}.$ 
Moreover, for any $\bm{p} \in \set{P}(\alpha),$ the provider's revenue is
identical. 
\end{theorem}

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